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IE 214: Operations Management. Lecture. 3. Forecasting. KAMAL. EXERCISE 4.3. Solution:. a) There is no pattern of any kind. EXERCISE 4.3. Solution:. EXERCISE 4.3. Solution:. d) The 3-year moving average method gives better results. EXERCISE 4.5. Solution:. (381+368+374)/3 = 374.33
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IE 214: Operations Management Lecture 3 Forecasting KAMAL
EXERCISE 4.3 Solution: a) There is no pattern of any kind
EXERCISE 4.3 Solution:
EXERCISE 4.3 Solution: d) The 3-year moving average method gives better results
EXERCISE 4.5 Solution: (381+368+374)/3 = 374.33 b) (381*0.1)+(368*0.3)+(374*0.6) = 372.9
EXERCISE 4.5 Solution: c) Ft = F t-1 + α(At-1 – Ft-1) ; α = 0.2
EXERCISE 4.8 Given:
EXERCISE 4.8 Solution: 91.3333 a) Ft = (96+88+90)/3 = 91.3333
EXERCISE 4.8 89 b) Ft = (88+90)/2 = 89
EXERCISE 4.8 c) MAD = ∑|At-Ft|/n = 13.5/5 = 2.7
EXERCISE 4.8 d) MSE = ∑(At-Ft)^2/n = 66.75/5 = 13.35
EXERCISE 4.8 e) MAPE = (∑|At-Ft|/At)*100/n = 0.149449*100/5 = 2.98897%
EXERCISE 4.19 Solution: Ft = F t-1 + α(At-1 – Ft-1) F3 = 48 = 50 + α(42 - 50) → -8α = -2 → α = 0.25 F5 = F4 + α(A4 – F4) = 50 + 0.25*(46 - 50) = 49
HW 4.1 4.4 4.14 4.45
EXERCISE 4.20 Given: α = 0.1 , ß = 0.2
EXERCISE 4.20 Solution: • Ft= α(At-1) + (1- α)(Ft-1 + Tt-1) • Tt= ß(Ft - Ft-1) + (1- ß)*Tt-1 • FIT = Ft + Tt
EXERCISE 4.20 65.5 0.1 65.6 • Ft= 0.1*(70) + (0.9)(65 + 0) = 65.5 • Tt= 0.2(65.5 - 65) + (0.8)*0 = 0.1 • FIT = 65.5 + 0.1 = 65.6
EXERCISE 4.21 • MSE= ∑(Error^2)/n = 12.7
EXERCISE 4.26 Given: a= y – b x Y = a + b x = 5 + (20)(5) = 105 ^
EXERCISE 4.26 Solution:
EXERCISE 4.26 Solution: b= (650)-(4)(2.5)(55)/[(30)-(4*(2.5^2))] = 100/5 = 20 a= 55 – (20*2.5) = 5 Y = a + b x = 5 + (20)(5) = 105 ^
HW 4.38 4.48
EXERCISE 4.30 Given: • 1000 radials produced each year. • Production planned to be 1200 radials next year.
EXERCISE 4.30 Solution: 1200 / 4 = 300 radials planned to be sold each season next year 300 * Is = Forecast for next year’s season
EXERCISE 4.30 Solution:
HW PROBLEMS 4.27 4.29